Answer:
a) 2.452
b) 1.256
Explanation:
Stress due to dead weight. = 14 Ksi
Stress due to fully loaded tractor-trailer = 45Ksi
ultimate tensile strength of beam = 76 Ksi
yield strength = 50 Ksi
endurance limit = 38 Ksi
Determine the safety factor for an infinite fatigue life
a) If mean stress on fatigue strength is ignored
β = ( 45 - 14 ) / 2
= 15.5 Ksi
hence FOS ( factor of safety ) = endurance limit / β
= 38 / 15.5 = 2.452
b) When mean stress on fatigue strength is considered
β2 = 45 + 14 / 2
= 29.5 Ksi
Ratio = β / β2 = 15.5 / 29.5 = 0.5254
Next step: applying Goodman method
Sa = [ ( 0.5254 * 38 *76 ) / ( 0.5254*76 + 38 ) ]
= 19.47 Ksi
hence the FOS ( factor of safety ) = Sa / β
= 19.47 / 15.5 = 1.256
Answer:
G = $37,805.65
Explanation:
I found this on another site:
475,000 = 25,000(P/A,10%,6) + G(P/G,10%,6)
475,000 = 25,000(4.3553) + G(9.6842)
9.6842G = 366,117.50
G = $37,805.65
Answer:
I would say false but I am not for sure
Answer:
3.03 INCHES
Explanation:
According to ASTM D198 ;
Modulus of rupture = ( M / I ) * y ----- ( 1 )
M ( bending moment ) = R * length of span / 2
= (120 * 10^3 ) * 48 / 2 = 288 * 10^4 Ib-in
I ( moment of inertia ) = bd^3 / 12
= ( 2 )*( d )^3 / 12 = 2d^3 / 12
b = 2 in , d = ?
length of span = 4 * 12 = 48 inches
R = P / 2 = 240 * 10^3 / 2 = 120 * 10^3 Ib
y ( centroid distance ) = d / 2 inches
back to equation ( 1 )
( M / I ) * y
940.3 ksi = ( 288 * 10^4 / 2d^3 / 12 ) * d / 2
= ( 288 * 10^4 * 12 ) / 2d^3 ) * d / 2
940300 = 34560000* d / 4d^3
4d^3 ( 940300 ) = 34560000 d ( divide both sides with d )
4d^2 = 34560000 / 940300
d^2 = 9.188 ∴ Value of d ≈ 3.03 in
